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$\mathbb{R}((t^{-1}))$ , the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property

Example

Let $K = \mathbb{R}((t^{-1}))$ be the field of formal Laurent series in $t^{-1}$ over $\mathbb{R}$ (The formal Laurent series $\mathbb{R}((t^{-1}))$ : support bounded below, valuation, leading coefficient), ordered by the sign of the leading coefficient ( $\mathbb{R}((t^{-1}))$ is an ordered field, ordered by the sign of the leading coefficient). This example assembles, in one place and against the five properties of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, what the field does and does not satisfy:

property	holds in $K$	reference
ordered field	yes	$\mathbb{R}((t^{-1}))$ is an ordered field, ordered by the sign of the leading coefficient
Archimedean	no	$\mathbb{R}((t^{-1}))$ is non-Archimedean, and the monomials $t^{-k}$ are cofinal below its positive elements
(CC) Cauchy completeness	yes	Every Cauchy sequence in $\mathbb{R}((t^{-1}))$ converges: $K$ is sequentially Cauchy complete
(NIP) nested intervals, shrinking	yes	$\mathbb{R}((t^{-1}))$ has the nested interval property for lengths tending to $0$
(LUB) least upper bound	no	$\mathbb{R}((t^{-1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum
(BW) Bolzano-Weierstrass	no	below
(MCT) monotone convergence	no	below

$K$ is therefore the witness for FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property, and a worked illustration of how far apart the two things called "completeness" can be. A concrete convergent Cauchy sequence is exhibited at the end.

Facts & Assumptions

Given: $K = \mathbb{R}((t^{-1}))$ , whose elements are the functions $\mathbb{Z} \to \mathbb{R}$ with support bounded below, with $t^{-a}$ the function taking the value $1$ at $a$ and $0$ elsewhere.

[L1]

$K$ is an ordered field, in which $f > 0$ exactly when $f \ne 0$ and its lowest-index nonzero coefficient is a positive real ( $\mathbb{R}((t^{-1}))$ is an ordered field, ordered by the sign of the leading coefficient, The formal Laurent series $\mathbb{R}((t^{-1}))$ : support bounded below, valuation, leading coefficient); and $(g-h)(j) = g(j) - h(j)$ ( $\mathbb{R}((t^{-1}))$ is a commutative ring: the product is a finite sum and both operations preserve support bounded below).

[L2]

$n \cdot 1_K < t$ for every natural $n$ , so $K$ is not Archimedean; $0_K < t^{-(k+1)} < t^{-k}$ ; for every $\varepsilon > 0$ in $K$ there is $k \in \mathbb{N}$ with $t^{-k} < \varepsilon$ ; and if $h(j) = 0$ for every $j \le k$ then $|h| < t^{-k}$ ( $\mathbb{R}((t^{-1}))$ is non-Archimedean, and the monomials $t^{-k}$ are cofinal below its positive elements, Archimedean ordered field).

[L3]

Every Cauchy sequence in $K$ converges in $K$ (Every Cauchy sequence in $\mathbb{R}((t^{-1}))$ converges: $K$ is sequentially Cauchy complete), which is (CC) (The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness, Sequences, convergence, Cauchyness, monotonicity, boundedness and closed intervals in an arbitrary ordered field).

[L4]

The set $A = \{\, n \cdot 1_K : n \in \mathbb{N}\,\}$ is nonempty, bounded above by $t$ , and has no least upper bound in $K$ , so (LUB) fails ( $\mathbb{R}((t^{-1}))$ does not have the least-upper-bound property; its canonical naturals have no supremum).

[L5]

Every nested sequence of closed intervals of $K$ whose lengths tend to $0$ in $K$ has exactly one common point, so (NIP) holds ( $\mathbb{R}((t^{-1}))$ has the nested interval property for lengths tending to $0$ ).

[L6]

(BW) implies the Archimedean property (Bolzano-Weierstrass alone forces the Archimedean property, so it needs no separate Archimedean hypothesis) and so does (MCT) (The monotone convergence property alone forces the Archimedean property, so it carries no separate Archimedean hypothesis); and the five properties are equivalent once the Archimedean property is supplied where needed (For an ordered field the five completeness properties are equivalent, provided the Archimedean property is assumed alongside nested intervals and Cauchy completeness).

Verification

technique · direct

1.1

$K$ is an ordered field.

1.2

$K$ is not Archimedean: $t$ exceeds every canonical natural.

1.3

$K$ has (CC).

1.4

$K$ does not have (LUB): the canonical naturals are nonempty and bounded above and have no supremum in $K$ .

1.5

$K$ has (NIP) in the shrinking form of The five completeness properties of an ordered field: least upper bound, monotone convergence, nested intervals, Bolzano-Weierstrass, and Cauchy completeness.

2.1

So $K$ is a Cauchy complete, non-Archimedean ordered field without the least-upper-bound property, which is what this example asserts, and it is the witness used in FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property and in FALSE: the nested interval property alone implies the least-upper-bound property.

step 1.1step 1.2step 1.3step 1.4step 1.5

2.2

$K$ has neither (BW) nor (MCT), since either would force $K$ to be Archimedean, which step 1.2 denies.

step 1.2L6

2.3

A concrete convergent Cauchy sequence: let $f^{(n)} := \sum_{k=0}^{n} t^{-k}$ , the function taking the value $1$ at each index $0 \le j \le n$ and $0$ elsewhere. For $n > m$ the difference $f^{(n)} - f^{(m)}$ vanishes at every index $j \le m$ , so $|f^{(n)} - f^{(m)}| < t^{-m}$ ; since the monomials $t^{-m}$ get below every positive element of $K$ , the sequence is Cauchy in $K$ . Its limit is the element $L$ with $L(j) = 1$ for $j \ge 0$ and $L(j) = 0$ for $j < 0$ , which lies in $K$ because its support is bounded below, and $f^{(n)} - L$ vanishes at every index $j \le n$ , so $|f^{(n)} - L| < t^{-n}$ and $f^{(n)} \to L$ in $K$ .

step 1.1step 1.3L1L2

3.1

The table of the Example is therefore established in every row, and $K$ separates Cauchy completeness from the least-upper-bound property.

step 2.1step 2.2step 2.3∎

Remarks

The one-line reason. Comparison in $K$ looks only at the first coefficient at which two elements differ, so $t$ is bigger than every real constant and $t^{-1}$ is smaller than every positive real constant. The naturals are therefore bounded, which kills (LUB), (BW) and (MCT) at a stroke. Meanwhile a Cauchy sequence in $K$ must have each of its coefficients eventually constant, and reading off those eventual values builds the limit; nothing about the naturals being cofinal is needed for that.
Why the limit above is not a sum. The notation $\sum_{k \ge 0} t^{-k}$ for $L$ is a name for a function, not an infinite sum (The formal Laurent series $\mathbb{R}((t^{-1}))$ : support bounded below, valuation, leading coefficient). What step 2.3 proves is a genuine limit in the order of $K$ , and it happens to agree with that notation; no notion of convergence is presupposed by the notation itself.
What this example does not give. It says nothing about $\mathbb{R}(t)$ , the other non-Archimedean field in this library (The rational function field $\mathbb{R}(t)$ ordered by the eventual sign is an ordered field, worked out), which is neither Cauchy complete nor nested-interval complete and cannot replace $K$ in any of these roles.
Uniqueness of the complete ordered field is untouched. $K$ is not a complete ordered field, so it is no counterexample to that uniqueness; it is a counterexample only to the habit of calling (CC) completeness.

Depends on

Used by

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Sources

Formal power series (Wikipedia) (standard reference, not scraped)
Completeness of the real numbers (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)

R((t−1))\mathbb{R}((t^{-1}))R((t−1)), the formal Laurent series field, is Cauchy complete, non-Archimedean, and lacks the least-upper-bound property